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Question Number 178522 by peter frank last updated on 17/Oct/22

Answered by MJS_new last updated on 17/Oct/22

$$y={\mathrm{e}}^x>0$$$$4y^4+y^3-6y^2+y+1=0$$$$\mathrm{this}\mathrm{has}\mathrm{no}‘‘{\mathrm{nice}}^{″}\mathrm{solution}$$$$y\approx 1.09783726106$$$$\Rightarrow x\approx .0933421181361$$

Commented by peter frank last updated on 18/Oct/22

$$\mathrm{more}\mathrm{clarification}\mathrm{please}\mathrm{step}3$$

Commented by MJS_new last updated on 18/Oct/22

$$4y^4+y^3-6y^2+y+1=0$$$$y^4+\frac{1}{4}{y}^3-\frac{3}{2}{y}^2+\frac{1}{4}y+\frac{1}{4}=0$$$$y=z-\frac{1}{16}$$$$z^4-\frac{195}{128}{z}^2+\frac{225}{512}z+\frac{14973}{65536}=0$$$$\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{finding}\alpha ,\beta ,\gamma$$$$(z^2-\alpha z-\beta )(z^2+\alpha z-\gamma )=0$$$$z^4-({\alpha }^2+\beta +\gamma )z^2-\alpha (\beta -\gamma )z+\beta \gamma =0$$$$\iff$$$$-({\alpha }^2+\beta +\gamma )=-\frac{195}{128}$$$$-\alpha (\beta -\gamma )=\frac{225}{512}$$$$\beta \gamma =\frac{14973}{65536}$$$$\mathrm{solve}\mathrm{the}{1}^{\mathrm{st}}\mathrm{for}\gamma ,\mathrm{then}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{for}\beta ,\mathrm{the}{3}^{\mathrm{rd}}$$$$\mathrm{becomes}$$$${\alpha }^6-\frac{195}{64}{\alpha }^4+\frac{5763}{4096}{\alpha }^2-\frac{50625}{262144}=0$$$$\alpha =\sqrt{u+\frac{65}{64}}$$$$u^3-\frac{27}{16}u-\frac{55}{64}=0$$$$\mathrm{this}\mathrm{has}\mathrm{no}‘‘{\mathrm{nice}}^{″}\mathrm{solution}\Rightarrow \mathrm{we}\mathrm{cannot}\mathrm{use}$$$$\mathrm{the}\mathrm{exact}\mathrm{solution}\mathrm{for}y$$

Commented by peter frank last updated on 22/Oct/22

$$\mathrm{thanks}$$

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